Problem: Let $h$ be a vector-valued function defined by $h(t)=\left(\dfrac2t,3e^{2t}\right)$. Find $h'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac2{t^2}+6e^{2t}$ (Choice B) B $\left(2\ln(t),\dfrac32e^{2t}\right)$ (Choice C) C $\left(-\dfrac2{t^2},6e^{2t}\right)$ (Choice D) D $\left(\dfrac2{t^2},3e^{t}\right)$
Solution: $h$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $h(t)=\left(\dfrac2t,3e^{2t}\right)$. Let's differentiate the first expression: $\dfrac{d}{dt}\left(\dfrac2t\right)=-\dfrac2{t^2}$ Let's differentiate the second expression: $\begin{aligned}\dfrac{d}{dt}(3e^{2t})&=2\cdot3e^{2t} \\\\&=6e^{2t}\end{aligned}$ Now let's put everything together: $\begin{aligned} h'(t)&=\left(\dfrac{d}{dt}\left(\dfrac2t\right),\dfrac{d}{dt}(3e^{2t})\right) \\\\ &=\left(-\dfrac2{t^2},6e^{2t}\right) \end{aligned}$ In conclusion, $h'(t)=\left(-\dfrac2{t^2},6e^{2t}\right)$.